The geometric mean is an unbiased estimator of the mean of which continuous distribution?

Question

Is there any continuous distribution expressible in closed form, whose mean is such that the geometric mean of the samples is an unbiased estimator for that mean?

Update: I just realized that my samples have to be positive (or else the geometric mean may not exist) so maybe continuous isn't the right word. How about a distribution which is zero for negative values of the random variable and is continuous for positive values. Something like a truncated distribution.

Answer 1

I believe you are asking what is, if any, the distribution of an r.v. $X$, such that, if we have an i.i.d. sample of size $n>1$ from that distribution, it will hold that

$$E[GM] = E\left[\left(\prod_{i=1}^n X_{i}\right)^{1/n}\right] = E(X)$$

Due to the i.i.d. assumption, we have

$$E\left[\left(\prod_{i=1}^n X_{i}\right)^{1/n}\right] = E\left(X_1^{1/n}\cdot ...\cdot X_n^{1/n}\right) = E\left (X_1^{1/n}\right)\cdot ...\cdot E\left(X_n^{1/n}\right) = \left[E\left(X^{1/n}\right)\right]^n$$

and so we are asking whether we can have

$$\left[E\left(X^{1/n}\right)\right]^n = E(X)$$

But by Jensen's inequality, and the fact that the power function is strictly convex for powers higher than unity, we have that, almost surely for a non-degenerate (non-constant) random variable,

$$\left[E\left(X^{1/n}\right)\right]^n < E\left[\left(X^{1/n}\right)\right]^n = E(X)$$

So no such distribution exists.

Regarding the mention of the log-normal distribution in a comment, what holds is that the geometric mean ($GM$) of the sample from a log-normal distribution is a biased but asymptotically consistent estimator of the median. This is because, for the lognormal distribution it holds that

$$E(X^s) = \exp\left\{s\mu + \frac {s^2\sigma^2}{2}\right \}$$

(where $\mu$ and $\sigma$ are the parameters of the underlying normal, not the mean and variance of the log-normal).

In our case, $s = 1/n$ so we get

$$E(GM) = \left[E\left(X^{1/n}\right)\right]^n = \left[\exp\left\{(\mu/n) + \frac {\sigma^2}{2n^2}\right \}\right]^n = \exp\left\{\mu + \frac {\sigma^2}{2n}\right \}$$

(which tells us that it is a biased estimator of the median). But

$$\lim \left[E\left(X^{1/n}\right)\right]^n = \lim \exp\left\{\mu + \frac {\sigma^2}{2n}\right \} = e^{\mu}$$

which is the median of the distribution. One can also show that the variance of the geometric mean of the sample converges to zero, and these two conditions are sufficient for this estimator to be asymptotically consistent - for the median,

$$GM \to_p e^{\mu}$$

Answer 2

This is a similar argument to Alecos's excellent answer since the arithmetic mean, geometric mean inequality is a consequence of Jensen's inequality.

• Let $A_n$ be the arithmetic mean: $A_n = \frac{1}{n} \sum_{i=1}^n X_i$

• Let $G_n$ be the geometric mean: $G_n = \left( \prod_{i=1} X_i \right) ^ \frac{1}{n}$

The arithmetic mean, geometric mean inequality states that $ A_n \geq G_n$ with equality if and only if every observation is equal: $X_1 = X_2 = \ldots = X_n$. (The AMGM inequality is a consequence of Jensen's inequality.)

Case 1: $X_1 = X_2 = \ldots = X_n$ almost surely

Then $ \operatorname{E}[G_n] = \operatorname{E}[A_n] = \operatorname{E}[X]$.

In some sense, this is an entirely degenerate case.

Case 2: $P(X_i \neq X_j) > 0$ for $i\neq j$

Then there's positive probability that the geometric mean is smaller than the arithmetic mean. Since for all outcomes $G_n \leq A_n$ and $\operatorname{E}[A_n] = \operatorname{E}[X]$, we then have $\operatorname{E}[G_n] < \operatorname{E}[X]$.